Question

If $y={\sin }^{-1}\sqrt{1-x^2}$, then $\frac{dy}{dx}=$

• A. $\frac{1}{\sqrt{1-x^2}}$
• B. $\frac{-1}{\sqrt{1-x^2}}$
• C. $\frac{1}{\sqrt{1+x^2}}$
• D. $\frac{-1}{\sqrt{1+x^2}}$

Answer 1

The correct option is B

$\frac{-1}{\sqrt{1-x^2}}$

Explanation for the correct option:

Find the value of $\frac{dy}{dx}$:

Given,

$y={\sin }^{-1}\sqrt{1-x^2}$

Now put,

$$\begin{gathered} x=\cos \theta \\ \theta ={\cos }^{-1}x \end{gathered}$$

Then,

$$\begin{gathered} y={\sin }^{-1}\sqrt{1-{\cos }^2\theta } \\ y={\sin }^{-1}\sqrt{{\sin }^2\theta }\mathbf{[}\mathbf{\because }\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathbf{c}\mathbf{o}{\mathbf{s}}^{\mathbf{2}}\mathbf{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{]} \\ y={\sin }^{-1}\left(\sin \theta \right) \\ y=\theta \\ y={\cos }^{-1}x \end{gathered}$$

Now differentiate with respect to $x$,

Then,

$\frac{dy}{dx}=\frac{-1}{\sqrt{1-x^2}}$

Hence, the correct option is B.